Latent heat is the heat per unit mass required for a phase change to occur. We may think about what happens when we add ice and cold soda to glass. Since the soda is warmer than zero degrees Celsius. So, we may expect that the heat from the slightly warmer soda to melt the ice in the glass. But, we have noticed that when we add ice to an already cold drink, only some of the ice melts, not all of it. Its reason is that the soda does not contain enough energy as heat to overcome the latent heat of fusion of the ice. This topic will explain the Latent heat of fusion formula with examples. Let us learn it!

**Latent Heat of Fusion**

**Concept **

When someone combines ice at zero degrees Celsius and cold soda at a temperature above zero degrees Celsius, then the heat from the soda will continue to melt the ice until both reach the temperature of equilibrium.

It must be remembered that during the phase change, the temperature of an object will remain constant despite the additional heat added or subtracted. As states of matter are gas, liquid, and solid. So, there are three different terms for latent heats are used.

The latent heat of fusion refers to the phase change between states of solid and liquid. Here, heat actually refers to the transfer of heat energy between the objects. Thus, the latent heat of fusion encompasses the process of adding heat to melt some solid.

**The formula for Latent heat of fusion:**

If m kg of the solid changes into the liquid at a constant temperature which is its melting point. Then the heat absorbed by it means the latent heat of fusion formula will be,

Q = \(m \times L_f\)

Also, if the temperature of some object varies from the lower temperature \(t_1\) to higher temperature \(t_2\). Then the heat gained or released by the material will be:

Q = \(m \times C \times \Delta t\)

i.e. \(Q = m \times C (t_2 – t_1)\)

Therefore, the total heat gained or released will be represented as,

Q = \(m \times L_f + m \times C \times \Delta t\)

Where,

\(L_f\) | specific latent heat of fusion |

Q | Heat |

m | Mass |

C | Specific heat |

## Solved Examples

Q.1: Find out the amount of water converted into ice, if 64500 calories of heat are extracted from the 100 g of steam at 100 degrees C. Given as the latent heat of ice and steam are 80 cal per gram and 540 cal per gram respectively.

Solution:

If the full steam is converted into the water at 100 degrees C, then the temperature of the water will fall from 100 degrees C to 10 degrees C. Also, a part of the water at 0 degree C for converting into the ice.

The heat gained to bring the steam at 100 C to water at 100 degrees C = \(m \times L_{steam}\)

= \(100 \times 540 cal \)

= 54000 cal.

Now, The heat gained to bring water at 100 degrees C into water at 0 degrees C,

= \(100 \times 1 \times (100-0) \)

= 10000 Cal

Thus, the total heat extracted = 54000 + 10000 = 64000 cal.

Remaining heat = 64500 – 64000 = 500 cal.

Therefore, amount of water converted into ice,

m = \(\frac {Q}{ L_{ice}} \)

= \(\frac{500}{ 80 } \)

= 6.25g.

The amount of water is 6.25 grams.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…